IMO 1984 SL 20

Origin: USA

Problem

Determine all pairs (a, b) of positive real numbers with a ̸= 1 such that loga b < loga+1(b + 1).

Solution

Define the set S = R+ ∖{1}. The given inequality is equivalent to ln b/ln a < ln (b + 1)/ln (a + 1). If b = 1, it is obvious that each a \inS satisfies this inequality. Suppose now that b is also in S. Let us define on S a function f(x) = ln (x + 1)/ln x. Since ln (x + 1) > ln x and 1/x > 1/x + 1 > 0, we have f ′(x) = ln x x+1 −ln (x+1) x ln2 x < 0 for all x. Hence f is always decreasing. We also note that f(x) < 0 for x < 1 and that f(x) > 0 for x > 1 (at x = 1 there is a discontinuity). Let us assume b > 1. From ln b/ln a < ln (b + 1)/ln (a + 1) we get f(b) > f(a). This holds for b > a or for a < 1. Now let us assume b < 1. This time we get f(b) < f(a). This holds for a < b or for a > 1. Hence all the solutions to loga b < loga+1(b + 1) are {b = 1, a \inS}, {a > b > 1}, {b > 1 > a}, {a < b < 1}, and {b < 1 < a}.